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bn vào /h7.net/hoi-dap/toan-6/so-sanh-a-3-10-1-3-9-1-va-b-3-9-1-3-8-1--faq205231.html
Trả lời:
A = \(\frac{3^{10}+1}{3^9+1}=\frac{3.3^9+1}{3.3^8+1}=\frac{3^9+1}{3^8+1}\)= B
_Học tốt bạn nha_
\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}=\frac{1+5\left(1 +5+5^2+...+5^8\right)}{1+5+5^2+...+5^8}=5+\frac{1}{1+5+5^2+...+5^8} \)
\(B=\frac{1+3+3^2+....+3^9}{1+3+3^2+....+3^8}=\frac{1+3\left(1+3+3^2+....+3^8\right)}{1+3+3^2+....+3^8}=3+\frac{1}{1+3+3^2+....+3^8}\)
\(=5+\frac{1}{1+3+3^2+....+3^8}-2\)
Có: \(\frac{1}{1+5+5^2+...+5^8}>0\) và \(\frac{1}{1+3+3^2+....+3^8}-2< 0\)
\(\Rightarrow A>B\)
A = 387420490 ; B = 1000000001
vậy B lớn hơn A
\(B=\left(\frac{1}{4}-1\right)\cdot\left(\frac{1}{9}-1\right).....\left(\frac{1}{81}-1\right)\cdot\left(\frac{1}{100}-1\right)\)
\(B=\frac{-3}{4}\cdot\frac{-8}{9}....\frac{-80}{81}\cdot\frac{-99}{100}\)
\(B=-\left(\frac{3}{4}\cdot\frac{8}{9}\cdot\cdot\cdot\cdot\frac{99}{100}\right)\)
\(B=-\left(\frac{3\cdot8\cdot15\cdot24\cdot....\cdot63\cdot80\cdot99}{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot\cdot\cdot9^2\cdot10^2}\right)\)
\(B=-\left(\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot\cdot\cdot\cdot8\cdot10\cdot9\cdot11}{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot\cdot9^2\cdot10^2}\right)\)
\(B=-\frac{11}{2\cdot10}\)
\(B=\frac{-11}{20}\)
\(B=\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right)...\left(\frac{1}{100}-1\right)\)
\(B=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-99}{10^2}\)
\(B=-\left(\frac{3}{2^2}.\frac{8}{3^2}...\frac{99}{10^2}\right)\)(có 9 thừa số, mỗi thừa số là âm nên kết quả là âm)
\(B=-\left(\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{9.11}{10.10}\right)\)
\(B=-\left(\frac{1.2...9}{2.3...10}.\frac{3.4...11}{2.3...10}\right)\)
\(B=-\left(\frac{1}{10}.\frac{11}{2}\right)\)
\(B=-\frac{11}{20}\)
\(M=\frac{1}{1.2}+\frac{2}{1.2.3}+.....+\frac{9}{1.2.3.....10}\)
\(M=\frac{2-1}{1.2}+\frac{3-1}{1.2.3}+....+\frac{10-1}{1.2......10}\)
\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{6}+....+\frac{10}{1.2.....10}-\frac{1}{1.2.....10}\)
\(M=1-\frac{1}{1.2.3......10}\)
\(M=1-\frac{1}{3628800}\)
Vì \(1=1\)mà \(\frac{1}{3628800}< 1\)nên \(1-\frac{1}{3628800}< 1\)
Vậy \(M< 1\)
\(\frac{10^9}{10^9-3}>\frac{10^9+2}{10^9-1}\)